p1 help needed!plz help me

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ashhadazam

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Well for this one the total opposite torques must be equal..
So: 900 x 0.20 = F x 1.20
U get F = 150 !!
 
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ashhadazam

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weight would be lifted if there is a minimum torque slightly grater than zero.. we assume them to be equal because even the slightest change would result in lifting of the weight ..
Well Q.10 is C
as the body was stationary the momentum is 0. find out the velocities.. M has velocity V while 2m has velocity 0.5V .. !! because this is how u r goign to get same momentums.. now find out the K.E's and get the ratio.. tell me if u arent able to get the anser :)
 
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ashhadazam

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In question 15, what u need to notice is that along with the height the mass is also reduced by 0.5 when the two water levels are equal. so P.E = 0.5m x g x 0.5h which gives u mgh/4 !!!
 
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ashhadazam

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Q.31 ) This one is easy.. the time is given just to confuse you .. You can use time if u use the equation E=VIT, but it is far easier if u use E=VQ .. divide E by V and u have your answer :)
 
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ashhadazam

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It indeed seems to be an interesting paper.. Keeping it for tuesday :D last mint preparations ;)
 
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nannes

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can any1 help me oct/nov 09 question 9-15 and 22...i knw a lot of questions but help meeeee
 
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musa

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q9 momentum before and after collision must be sAME
MOMENTUM BEFORE COLLISION IS 2MU+(-MU)=MU
SO ONLY IN A THE TOTAL MOMENTUM IS (5/3MU)+(-MU/3)=MU
 
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musa

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Q10
CALCULATE ACCELARATION BY a=F/M =60/30=2
so use equation v=u+at take accelaration negative as its slowing down
3+(-2*0.5)=2
so C

Q11
its the point through which the weight may be considered to act
 
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